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# A sum becomes 8000 in 3 years and 10000 in 6 years at C.I. Find the sum.  Hint: We can use the formula for finding the amount $=p{{\left( 1+\dfrac{R}{100} \right)}^{n}}$ where $p$ is principal, $R$ is the rate of interest and $n$ is time.

We have been given the question that a sum becomes 8000 in 3 years and 10000 in 6 years at C.I. Here, we have to find the principal.
Before proceeding with the solution, we must know that the initial money is the principal and after adding interest to the principal, we get the final amount.
According to the question, we have to find the sum, that means we need to find the principal. Therefore, we have to use the formula for amount = $p{{\left( 1+\dfrac{R}{100} \right)}^{n}}$.
We are supposed to take the sum as the principal. Let the principal be $p$.
We have the amount$=8000$and the time period as 3 years. Therefore, we can substitute the values in the formula and we will get,
$\therefore 8000=p{{\left( 1+\dfrac{R}{100} \right)}^{3}}..........(i)$
Again, we have the amount$=10000$and the time period as 6 years. Therefore, we can substitute the values in the formula and we will get,
$\therefore 10000=p{{\left( 1+\dfrac{R}{100} \right)}^{6}}...........(ii)$
We can write $p{{\left( 1+\dfrac{R}{100} \right)}^{6}}$ as$p{{\left( 1+\dfrac{R}{100} \right)}^{3}}{{\left( 1+\dfrac{R}{100} \right)}^{3}}$.

Now, equation (ii) will be converted as:
$\Rightarrow 10000=p{{\left( 1+\dfrac{R}{100} \right)}^{3}}{{\left( 1+\dfrac{R}{100} \right)}^{3}}.....(iii)$
We can substitute the value of $p{{\left( 1+\dfrac{R}{100} \right)}^{3}}$ from equation (i) in equation (iii) and we will get,
$\Rightarrow 10000=8000{{\left( 1+\dfrac{R}{100} \right)}^{3}}$
$\Rightarrow \dfrac{10000}{8000}={{\left( 1+\dfrac{R}{100} \right)}^{3}}................(iv)$
No, we can substitute the value of ${{\left( 1+\dfrac{R}{100} \right)}^{3}}$ from equation (iv) in equation (i) and we will get,
$\Rightarrow 8000=p\left( \dfrac{10000}{8000} \right)$
By cross multiplying, we will get,
$\Rightarrow p=\dfrac{64000000}{10000}$
$\Rightarrow p=6400$
$\therefore$ Principal$=6400$
Hence, the sum is $6400$.

Note: Do not confuse compound interest with simple interest otherwise you will be messed up with your answer. Always remember that sum is taken as principal in both compound interest and simple interest, the final money will be taken as the amount. In these types of questions, if we are not given the rate of interest, then we need not find the rate. It is because the rate will get cancelled while taking the ratios of the formula for different time periods. So, do not get confused if the rate of interest is not given.
Since in formula, we have the amount $p{{\left( 1+\dfrac{R}{100} \right)}^{n}}$. So in case you are provided with time more than 3 years then, try to get the values from other equations. Do not split the powers.
Like, for example, do not do like:
${{\left( 1+\dfrac{R}{100} \right)}^{4}}=\left( 1+\dfrac{R}{100} \right)\left( 1+\dfrac{R}{100} \right)\left( 1+\dfrac{R}{100} \right)\left( 1+\dfrac{R}{100} \right)$
View Notes
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